You are given two lists A and

$$

$$both containing the numbers

$1$

dotsn in some

order. Your task is to count how many of the numbers

$1$

dotsn occur earlier on the

list A than on the list

$$

$.$

In this task,

$$

$$can be large and an efficient algorithm is required. The time

complexity should be

$$

$($

$$

$)$

$.$

In a file

twolists.py

$,$

$$implement a function count that returns the desired count.

def count

$($

a

$,$

$$b

$)$

:

# TODO

if

$=$

$=$

$"$

$"$

$"$

:

print

$($

count

$($

$[$

$2$

$,$

$3$

$,$

$4$

$,$

$1$

$]$

$,$

$[$

$1$

$,$

$2$

$,$

$3$

$,$

$4$

$]$

$)$

$)$

$$#

$3$

print

$($

count

$($

$[$

$1$

$,$

$2$

$,$

$3$

$,$

$4$

$]$

$,$

$[$

$1$

$,$

$2$

$,$

$3$

$,$

$4$

$]$

$)$

$$#

$0$

print

$($

count

$($

$[$

$4$

$,$

$7$

$,$

$3$

$,$

$1$

$,$

$6$

$,$

$2$

$,$

$5$

$]$

$,$

$[$

$5$

$,$

$6$

$,$

$1$

$,$

$2$

$,$

$4$

$,$

$3$

$,$

$7$

$]$

$)$

$$#

$3$

print

$($

count

$($

$[$

$5$

$,$

$4$

$,$

$9$

$,$

$1$

$,$

$8$

$,$

$3$

$,$

$2$

$,$

$6$

$,$

$7$

$]$

$,$

$[$

$6$

$,$

$2$

$,$

$8$

$,$

$4$

$,$

$9$

$,$

$1$

$,$

$5$

$,$

$7$

$,$

$3$

$]$

$)$

Explanation: In the first test, the numbers

$2$

$,$

$3$

$$and

$4$

$$occur earlier on the list

$$

than on the list

$$

$.$